Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $a = \dfrac{8r - 16}{4r + 36} \div \dfrac{r^2 - 2r}{r^2 + 7r - 18} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $a = \dfrac{8r - 16}{4r + 36} \times \dfrac{r^2 + 7r - 18}{r^2 - 2r} $ First factor the quadratic. $a = \dfrac{8r - 16}{4r + 36} \times \dfrac{(r + 9)(r - 2)}{r^2 - 2r} $ Then factor out any other terms. $a = \dfrac{8(r - 2)}{4(r + 9)} \times \dfrac{(r + 9)(r - 2)}{r(r - 2)} $ Then multiply the two numerators and multiply the two denominators. $a = \dfrac{ 8(r - 2) \times (r + 9)(r - 2) } { 4(r + 9) \times r(r - 2) } $ $a = \dfrac{ 8(r - 2)(r + 9)(r - 2)}{ 4r(r + 9)(r - 2)} $ Notice that $(r - 2)$ and $(r + 9)$ appear in both the numerator and denominator so we can cancel them. $a = \dfrac{ 8\cancel{(r - 2)}(r + 9)(r - 2)}{ 4r\cancel{(r + 9)}(r - 2)} $ We are dividing by $r + 9$ , so $r + 9 \neq 0$ Therefore, $r \neq -9$ $a = \dfrac{ 8\cancel{(r - 2)}\cancel{(r + 9)}(r - 2)}{ 4r\cancel{(r + 9)}\cancel{(r - 2)}} $ We are dividing by $r - 2$ , so $r - 2 \neq 0$ Therefore, $r \neq 2$ $a = \dfrac{8(r - 2)}{4r} $ $a = \dfrac{2(r - 2)}{r} ; \space r \neq -9 ; \space r \neq 2 $